l-Hospital's rule problem?

[baconhawk]

I have figured out that this is an indeterminate form of 0/0 but I am having a hard time taking all the derivatives.
The problem is the limit as x approaches zero of (3x-sin3x)/(3x-tan3x)

 

[soroban]

Hello, baconhawk!

I have figured out that this is an indeterminate form of 0/0,
but I am having a hard time taking all the derivatives. . Why?

. . $\displaystyle \displaystyle \lim_{x\to0} \frac{3x-\sin3x}{3x-\tan3x}$

You don't know the derivative of $\displaystyle 3x\,?$

You don't know the derivative of $\displaystyle \sin3x\,?$

You don't know the derivative of $\displaystyle \tan3x\,?$

Exactly where is your difficulty?

 

[baconhawk]

Well I know that using l'Hospital's Rule it then becomes the limit as x approaches 0 of (3-3cos3x)/(3-3sec^2(3x)) which is still an indeterminate form. I guess that my trouble is in getting to a form that does not become an indeterminate when I take the limit.
x→0

 

[Bob Brown MSEE]

(3x-sin3x)/(3x-tan3x)

Let t=3x

(t-sint)/(t-tant)

(t-(t-t³/6+...))/(t-(t+t³/3+....))

(-t³/6+...))/(+t³/3+....))

Higher Terms don't matter (or differentiate these twice to prove it)

Answer = -1/2

 

[Bob Brown MSEE]

Trick Question

The multiple derivatives still give 0/0.
The Maclaurin series has non zero second derivatives.

 

[Deleted member 4993]

The multiple derivatives still give 0/0

Not really....

(T-sinT)/(T-tanT) → (1 - cosT)/(1-sec²T) = (cosT-1)/tan²T → (-sinT)/(2tanT*sec²T) = -1/2 * (cos³T) → -1/2